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hi i could really use some help with this pre lab, and will award all the point

ID: 783670 • Letter: H

Question

hi i could really use some help with this pre lab, and will award all the point to the person who helps the most.


https://www.dropbox.com/s/pauj005wkcby0m1/probsetH2O2.pdf


i have done most of the work i just need help solving parts: C, E, F, I.


i found that:

A) Molarity of MnO4- solution ...0.0264 M

B) Molarity of HOOH in the unknown...1.32 M

C) Partial pressure of O2 collected.........??? mmHg

D) Volume of O2 gas collected...53.3 mL

E) Milimoles of O2 generated.........??? mmol

F) Molarity of HOOH in unknown.........??? M

G) Absorbance of the STANDARD...0.324

H) Absorbance of the UNKNOWN...0.404

I) Molarity of original UNKNOWN.........??? M


Thanks for your help in advance!

Explanation / Answer

E)


Here's what I understand from Method 2:


- You have a test tube and in it are MnO2 and 4.90 mL of your unknown

- There's a gas collection flask that is upside down

- Both the test tube and collection flask are in water


I am not really understanding what is meant when it says "the weight of the collection flask with remaining water" because the collection flask has no water to begin with.


Here's the calculation that I came up with:


Mass of water = 263.62 g - 210.52 g = 53.1 g


Moles of water = 53.1 g x 1 mol / 18.02 g = 2.95 moles


Use the mole to mole ratio of H2O to O2 from the balanced chemical equation to solve for the moles of O2:


2.95 moles of H2O x 1 mole of O2 / 2 moles of H2O = 1.48 moles


mmoles of O2 = 1.48 moles x 1,000 mmoles / 1 mole = 1,480 mmoles


C)


Use the ideal gas law, PV = nRT, to solve for the total moles:


P = 766.0 mmHg = 1.01 atm


V = 4.90 mL = 0.0049 L


R = 0.0821 atm*L/mol*K (gas constant)


T = 21 deg C = 294.15 K


n = ?


PV = nRT


n = PV / RT


n = (1.01 atm) 0.0049 L) / (0.0821 atm*L/mol*K)( 294.15 K) = 17.7 moles


Partial pressure of O2 = mole fraction of O2 x total pressure


Mole fraction of O2 = 1.48 moles / 17.7 moles = 0.0836


Partial pressure of O2 = 0.0836 x 1.01 atm = 0.0844 atm


These are the two problems I know how to solve. The other two I did know from before, but forgot.


Hope this helps! :)